The edge of a cube is increasing at a rate of 7 cm/s. The rate of change of area of the cube when edge of the cube is 3 cm is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(252\, \mathrm{cm^2/s}\)
The area of a cube is its total surface area, made up of six square faces. With edge x cm:
\(S = 6x^2\)
Differentiating with respect to time t (chain rule):
\( \displaystyle \frac{dS}{dt} = 12x\frac{dx}{dt} \)
Putting \(x = 3\) and \(\frac{dx}{dt} = 7\):
\( \displaystyle \frac{dS}{dt} = 12 \times 3 \times 7 = 252 \)
So the area increases at 252 cm2/s.
Why the other options are incorrect
Option (1) \(42\, \mathrm{cm^2/s}\) — Incorrect. This is the rate for one face only: \(\frac{d}{dt}(x^2) = 2 \times 3 \times 7 = 42\).
Option (2) \(168\, \mathrm{cm^2/s}\) — Incorrect. This is the rate for the lateral surface area \(4x^2\) (four faces): \(8 \times 3 \times 7 = 168\).
Option (3) \(84\, \mathrm{cm^2/s}\) — Incorrect. The total surface area \(6x^2\) changes at \(12 \times 3 \times 7 = 252\) cm2/s, not 84.